Question

In the given figure, ABCD is a parallelogram. Prove
that : AB = 2 BC.​

Answer 1

Answer:

Step-by-step explanation:

Let draw a line parallel to both lines BC and AD  

now in \triangle AEO\quad and\quad \triangle ADE\\ \dfrac { AE }{ AO } =\dfrac { AE }{ AD }△AEOand△ADE

AO

AE

​  

=  

AD

AE

​  

 (by Thales's theorm)

\\ \Rightarrow { AO }={ AD }={ BC }

⇒AO=AD=BC

similarly in\\ \triangle CBE\quad and\quad \triangle OBC\\ \dfrac { EB }{ BC } =\dfrac { EB }{ OB }

△CBEand△OBC

BC

EB

​  

=  

OB

EB

​  

 (by Thales's theorm)

\\ \Rightarrow { OB }={ BC }\\ { OA }+{ OB }={ BC }+{ BC }\\ \Rightarrow { AB }={ 2BC }

⇒OB=BC

OA+OB=BC+BC

⇒AB=2BC

Proved

Answer 2

Step-by-step explanation:

IN ABCD a parallelogram,

ABCD IS RECTANGLE, ABE IS A TRIANGLE

AB=CD (OPPOSITE SIDES IN A PARALLELOGRAM)

AD= BC (SAME REASON)

THEREFORE, AB=2×BC (OR AD)

HOPE IT HELPS YOU BRO